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  1. I was wondering which one is more general, metric linear spaces or locally convex topological vector spaces?

    Is a metric linear space a locally convex topological vector space? Vice versa?

  2. In terms of the number of books and websites that have mentioning, it seems like metric linear spaces are less popular than locally convex topological vector spaces, although both are topological vector spaces and normed vector spaces are both, doesn't it? Why is that?

    For example, there is no Wiki article for metric linear spaces, but there is one for locally convex topological vector spaces.

  3. As to metric linear spaces, the definition I saw from this book says a metric linear space is a vector space with a metric, such that addition and scalar multiplication are both continuous wrt the metric. That the metric is translation invariant is stronger than that definition. Isn't it?

Thanks and regards!

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I've seen the additional requirement that addition in a metric linear space be uniformly continuous. Presumably because such a space is also an abelian topological group and thus has a uniformity that makes addition uniformly continuous and we don't want our metric to not play well with the uniform structure. – kahen Oct 24 '12 at 22:42
One of the prototypical examples of why we want local convexity is $L_p$ spaces for $0<p<1$. The space $L_p(X,\mu)$ turns out to be a complete metric linear space under the metric $d_p(f,g) := \Bigl(\int |f-g|^p \,d\mu\Bigr)^{1/p}$. But $f \mapsto d_p(f,0)$ is not a norm on $L_p(X,\mu)$. And $L_p([0,1],$ Lebesgue measure$)$ even has trivial continuous dual. See for example Meise & Vogt - Introduction to Functional Analysis. – kahen Oct 24 '12 at 22:43
up vote 1 down vote accepted

Reading your question and comments, it seems that there are actually three concepts you are asking about:

  • metric vector spaces (i.e. vector spaces equipped with a metric such that the induced topology on the vector space makes scalar multiplication and addition continuous, or, what is the same, makes the vector space a topological vector space)

  • metrisable vector spaces (i.e. topological vector spaces whose topology can be defined by a metric; but no particular choice of metric is given as part of the structure)

  • locally convex topological vector spaces (i.e. topological vector spaces such that each point has a neighbourhood basis of convex sets)

Certainly one can find metric vector spaces in which the metric is not translation invariant. But one can also find an equivalent metric (i.e. one inducing the same topology) which is transation invariant. Thus, if one is given a metrisable vector spaces, one can always assume that its topology arises from a translation invariant meteric.

Finally, as the other answers have indicated, a metrisable vector space does not have to be locally convex, and nor does a locally convex topological vector space have to be metrisable. There is a name for topological vector spaces which are metrisable, locally convex, and complete: they are called Frechet spaces.

As to which one is more natural to consider, this depends on what you are using them for. As indicated by Norbert, one reason that people require the locally convex condition is that this is the condition that is required for the Hahn--Banach theorem to be true; a topological vector space that is not locally convex need not have any non-zero continuous linear forms defined on it.

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A locally convex topological vector space need not be metrizable; the second-last example mentioned here is not metrizable. The examples listed here are metrizable (in some cases even normable) vector spaces that are not locally convex.

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+1 Thanks! How do you think about the two possibly different definitions of metric linear spaces? And good morning! (I perhaps need to go back to sleep again.) – Tim Feb 26 '12 at 11:22
@Tim: You’re right: continuity of addition and scalar multiplication does not imply translation invariance. The metric $$d(x,y)=\frac{|x-y|}{1+|x-y|}$$ on $\mathbb{R}$ generates the usual topology, so addition and multiplication are continuous, but it’s not translation invariant. (I’m about to go to bed myself: I do most of my sleeping during the day.) – Brian M. Scott Feb 26 '12 at 11:29
@BrianM.Scott how is that not translation-invariant? $$d(x+z,y+z) = \frac{|(x+z)-(y+z)|}{1+|(x+z)-(y+z)|} = \frac{|x-y|}{1+|x-y|} = d(x,y).$$ But of course it's not locally convex (which I presume is what you meant). – kahen Oct 24 '12 at 22:24
@kahen: Or I may have been falling asleep when I wrote that; at this remove in time I honestly don’t remember what I had in mind. – Brian M. Scott Oct 24 '12 at 23:13

You can endow linear space with different metrics, which in general won't fit the linear structure of the space. The most natural requirement for the metric on the linear space is to be translation invariant: $$ d(x+z,y+z)=d(x,y) $$ But not every translation invariant metric gives a locally convex topology. You can consider the space of measurable functions on the interval $[0,1]$ with metric defined by $$ d(f,g)=\int_{[0,1]}\frac{|f(x)-g(x)|}{1+|f(x)-g(x)|}d\mu (x) $$ It is not locally convex. Local convexity of the linear space is a very good property which guarantees existence of sufficient family of continuous linear functionals on this linear space.

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+1 Thanks! (1) The definition I saw from this book says a metric linear space is a vector space with a metric, such that addition and scalar multiplication are both continuous wrt the metric. The one you mentioned, i.e. the metric is translation invariant is more stronger than that definition. Isn't it? (2) Do metric linear spaces (with possibly two different definitions) and locally convex TVSes imply each other? – Tim Feb 26 '12 at 11:13
You should take a look at the book of W. Rudins Functional Analysis. It contains a lot of general interesting facts about topological vector spaces. – Norbert Feb 26 '12 at 11:33
Thank you! Does it say about metric linear spaces? – Tim Feb 26 '12 at 11:35
It say about Frechet spaces, which are seems to be the same as yours metric linear spaces. – Norbert Feb 26 '12 at 11:50

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